I noticed that pixel and maverick also aluded to electronic devices, which is a different issue for which current is also important, so I just wanted bring that back into the conversation as well. Modern electronics use electricity in a sophisticated way. That is, electric and magnetic fields are important design considerations, as well as the motion/mobility of charge carries. For instance, the operation of a BJT (a still quite common type of transistor) is most straightforwardly modeled in terms of the current that flows into and out of the connections.
user111_23 said:
is resistance a factor in the heating of wires?
Yes, mathematically. The simple version of electric heating is RI
2. But it is the current that brings the energy into the wires; the resistance is just a parameter. As has been pointed out, this can also be expressed as VI for resistive/ohmic devices, since V, I and R are related. However, if the device is highly reactive, then only part of that voltage goes into the heating, and RI
2 is the more direct calculation, where R is the resistive part of the impedance of the device. (I hope that I am not confusing you more. Look up "power factor".)
user111_23 said:
how does current determine how efficient a circuit is?
Consider the power lines that run from the power station to your house. These lines transport a lot of electrical power; they need to supply your house with all of the needed electricity, as well as hundreds of other houses.
I will simplify and approximate the situation a bit for the sake of clarity:
In order for you to get, say 1 kW at 100 V, that would require 1 kW / 100 V = 10 A to be transmitted down the power line. If the electric company actually did that over, say 10 km, and suppose the resistance of the power line was 1 Ω/km, then the power loss in the line would be
( 1 Ω/km * 10 km ) * ( 10 A )
2 = 1 kW.
That is, the 10 A distribution would result in a 50% efficiency. However, the electric company does not do this. Instead, they use a transformer to transfer the power at a much lower current, say 1 A for this direct residential application. Then, the power loss in the line would be
( 1 Ω/km * 10 km ) * ( 1 A )
2 = 10 W.
That is, the 1 A distribution would result in a 99% efficiency. I have ignored other loss mechanisms, such as the loss in the transformers, but the basic idea is that lower current makes the transmission more efficient.